ATP & Bioenergetics

I. Introduction

II.  ATP

A. What is it?

B. How does it function?

Some example data:

Tissue ATP ADP AMP
Rat liver 4.7 1.5 0.6
Rat heart 13.3 2.6 0.43
Blowfly muscle 14.0 3.0 0.25

The sum of the concentrations of ATP, ADP and AMP will remain relatively constant in a cell, but depending on the metabolic rate, the percentage of ATP will be higher.

C. Energy Charge

The relative amount of high energy forms of ATP (ATP, ADP) can be calculated using the following formula:

EC = 1/2([ADP] + 2[ATP]/[AMP] + [ADP] + [ATP])

Note that if all adenosine phosphates are ATP, EC = 1.0; if all AMP, EC = 0; and if all ADP or ATP = AMP, EC = 0.5

D. Free energy and ATP

How does the energy in ATP specifically get utilized to power reactions in metabolism?

First Law: In any process, the total energy of the systems and the surroundings remains constant; energy is not created or destroyed.  Energy may be transformed from one form to another

Second Law: In any process, the entropy of the system and the surroundings increases. Entropy is often thought of as disorder or randomness.

 

Free Energy

DG =DH - TDS

where, G = free energy; H = enthalpy (total energy contained in chemical compound); S = entropy; T = absolute temperature

A + B <=>C + D

DG = DG0 + RT ln ([C][D]/[A][B])


"The Chemists Free Energy"

"The Biologists Free Energy"

R= gas constant (8.31 X 10-3 kJ/mole 0K)

T= absolute temperature

Calculate the standard free energy for a chemical reaction assuming the following conditions:

Let the reaction go to equilibrium where no net chemical change is occurring and the free energy change for the system is at a minimum (DG = 0)

Example reaction:

Glucose-1-phosphate <=> Glucose-6-phosphate

This reaction occurs in glycolysis and is catalyzed by the enzyme phosphoglucomutase. Assume you start with 0.020 M Glucose-1-phosphate, add the enzyme (along with other appropriate reaction conditions; 25 0C) and allow the reaction to proceed to equilibrium. The final equilibrium mixture was found to have:

0.001M Glucose-1-phosphate

0.019M Glucose-6-phosphate

The Keq = Glucose-6-phosphate/ Glucose-1-phosphate = 0.019/0.001 = 19

DG0= - RT ln Keq

= - 8.31 X 10-3 kJ/mole 0K x 298 x ln 19

= -7.29 kJ/mole

Note that the value for DG0 is negative. What does this mean? What type of reaction is this?

From an energetic standpoint, what does this mean?

Is it possible for these reactions to occur in the cell anyway?

E. The Coupling of Chemical Reactions; ATP

For example:

Glucose-1-phosphate <=> fructose-6-phosphate

This reaction is catalyzed by an isomerase and has a DG0 = +1.86 kJ/mole; this means it will not proceed forward spontaneously under standard conditions (1M concentrations, etc.).

The Solution:

fructose-6-phosphate + ATP => fructose-1,6-diphosphate + ADP

DG0 = -15.8 kJ/mole

Direct coupling by sharing a common enzyme

glucose + phosphate => glucose-6-phosphate (DG0 = 15.4 kJ/mole)

ATP + H2O =>  ADP + phosphate (DG0 = -34 kJ/mole)

The synthesis of glucose-6-phosphate would never occur under cellular conditions unless the reactions are added together:

glucose + ATP => glucose-6-phosphate + ADP

DG0 = -34 + 15.4 = -18.6 kJ/mole; reaction proceeds spontaneously in the direction written.